How to figure out Laurent series expansion for $z^2 \sin(1/(z+i))$?

Question

I am having trouble figuring out the Laurent series expansion for $z^2 \sin\dfrac{1}{z+i}$.

I would really appreciate any help.

Answer 1

firstly, we have $z^2=(z+i-i)^2=(z+i)^2-2(z+i)-1$ Then, $sin({1\over {z+i}})$=$1\over {z+i}$-$1\over {6{(z+i)}^3}$+${1\over {120{(z+i)}^5}}-...$ So you can just figure out the product of those two laurent series, then you will get answer.

Answer 2

Assuming you want to expand it around $-i$:

First expand $\sin \frac{1}{z + i}$ around $-i$ by expanding $\sin$ around $0$, then expand $z²$ around $-i$ and lastly multiply those expansions.

This will work, because $\sin$ is an entire function so its expansion around $0$ is valid for all of $ℂ$.